Question 610924
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^3\ +\ 3x^2\ -\ 8x\ -\ 24]

Rearrange

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^3\ -\ 8x\ +\ 3x^2\ -\ 24]

Factor *[tex \Large x] from the first two terms and factor 3 from the last two terms:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x(x^2\ -\ 8)\ +\ 3(x^2\ -\ 8)]

Note the common binomial factor, factor it out:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ 3)(x^2\ -\ 8)]

Now, depending on whether you are required to factor over the rationals (if so, you are done) or over the reals (if so, continue)

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 8]

looks very much like the difference of two squares, except that 8 isn't a square.  Or is it...?

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sqrt{8}\ =\ 2\sqrt{2}]

So:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 8\ =\ \left(x\ +\ 2\sqrt{2}\right)\left(x\ -\ 2\sqrt{2}\right)]

And your complete factorization, over the reals, is:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ 3) \left(x\ +\ 2\sqrt{2}\right)\left(x\ -\ 2\sqrt{2}\right)]

John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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